How do you differentiate #f(x)= ( 2 - xsecx )/ (x -3) # using the quotient rule?

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Answer 1 · 2017-03-23T03:22:53

#d/dxf(x) = ((-xsec(x)tan(x)-sec(x))(x-3) - (2-xsecx))/(x-3)^2#

(For this problem, I am using #h(x)# where #f(x)# is traditionally used, to avoid ambiguity since the original function is named #f(x)#)

The quotient rule states that:

#d/(dx)f(x) = d/(dx) (h(x))/g(x) = (h'(x)g(x)-h(x)g'(x))/g(x)^2#

In this case, we can say that:

#h(x) = 2-xsecx#
#g(x) = x-3#

Therefore, using standard differentiation rules:

#h'(x) = -xsec(x)tan(x)-sec(x)#
#g'(x) = 1#

Now we can plug these values into the quotient rule formula:

#d/dxf(x) = ((-xsec(x)tan(x)-sec(x))(x-3) - (2-xsecx))/(x-3)^2#

This is the final form of the derivative, but it can be expanded and simplified if necessary.

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